# Find the sum of all the terms in the n-th row of the given series

Find the sum of all the terms in the nth row of the series given below.

```               1  2
3  4  5  6
7  8  9 10 11 12
13 14 15 16 17 18 19 20
..........................
............................
(so on)
```

Examples:

```Input : n = 2
Output : 18
terms in 2nd row and their sum
sum = (3 + 4 + 5 + 6) = 18

Input : n = 4
Output : 132
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum.
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.

Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula:

``` Sum(n) = n * (2 * n2 + 1)
```

The proof for the formula is given below:

Prerequisite:

1. Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as:
```   Sum = (n * [2*a + (n-1)*d]) / 2
```
2. Sum of 1st n natural numbers is given as:
```   Sum = (n * (n + 1)) / 2
```

Proof:

```
Let the number of terms from the beginning
till the end of the nth row be p.
Here p = 2 + 4 + 6 + .....n terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n terms of the AP series, we get,

p = n * (n + 1)

Similarly, let the number of terms from the
beginning till the end of the (n-1)th row be q.
Here q = 2 + 4 + 6 + .....n-1 terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n-1 terms of the AP series, we get,

q = n * (n - 1)

Now,
Sum of all the terms in the nth row
= sum of 1st p natural numbers -
sum of 1st q natural numbers

= (p * (p + 1)) / 2 - (q * (q + 1)) / 2

Substituting the values of p and q and then solving
the equation, we will get,

Sum of all the terms in the nth row = n * (2 * n2 + 1)

```

## C++

 `// C++ implementation to find the sum of all the ` `// terms in the nth row of the given series ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// function to find the required sum ` `int` `sumOfTermsInNthRow(``int` `n) ` `{ ` `    ``// sum = n * (2 * n^2 + 1) ` `    ``int` `sum = n * (2 * ``pow``(n, 2) + 1); ` `    ``return` `sum; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << ``"Sum of all the terms in nth row = "` `         ``<< sumOfTermsInNthRow(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the sum of all the ` `// terms in the nth row of the given series ` ` `  `import` `static` `java.lang.Math.pow; ` ` `  `class` `Test { ` `    ``// method to find the required sum ` `    ``static` `int` `sumOfTermsInNthRow(``int` `n) ` `    ``{ ` `        ``// sum = n * (2 * n^2 + 1) ` `        ``int` `sum = (``int``)(n * (``2` `* pow(n, ``2``) + ``1``)); ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(``"Sum of all the terms in nth row = "` `                           ``+ sumOfTermsInNthRow(n)); ` `    ``} ` `} `

## Python3

 `# Python 3 implementation to find  ` `# the sum of all the terms in the ` `# nth row of the given series ` `from` `math ``import` `pow` ` `  `# function to find the required sum ` `def` `sumOfTermsInNthRow(n): ` `     `  `    ``# sum = n * (2 * n^2 + 1) ` `    ``sum` `=` `n ``*` `(``2` `*` `pow``(n, ``2``) ``+` `1``) ` `    ``return` `sum` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `4` `    ``print``(``"Sum of all the terms in nth row ="``,  ` `                   ``int``(sumOfTermsInNthRow(n))) ` ` `  `# This code is contributed ` `# by Surendra_Gangwar `

## C#

 `// C# implementation to find the sum of all the ` `// terms in the nth row of the given series ` `using` `System; ` ` `  `class` `Test { ` `    ``// method to find the required sum ` `    ``static` `int` `sumOfTermsInNthRow(``int` `n) ` `    ``{ ` `        ``// sum = n * (2 * n^2 + 1) ` `        ``int` `sum = (``int``)(n * (2 * Math.Pow(n, 2) + 1)); ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4; ` `        ``Console.Write(``"Sum of all the terms in nth row = "` `                      ``+ sumOfTermsInNthRow(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```Sum of all the terms in nth row = 132
```

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Improved By : jit_t, SURENDRA_GANGWAR

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